How do you factor completely $x^6-1$ ?

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Answer 1 · 2016-05-13T11:58:19

$(x-1)(x+1)(x^2-x+1)(x^2+x+1)$
$=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)(x+1/2-i sqrt3/2)(x+1/2+ i sqrt3/2)$

The answer in the given form can be obtained directly by observing that it is a cubic in $x^2$ .

$x^6-1=(x^2)^3-1^3$

$=(x^2-1)((x^2)^2+x^2+!)$

$=(x-1)(x+1)(x^4+x^2+1)$

$=(x-1)(x+1)(x^2-x+1)(x^2-x+1)$

$=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)$

$(x+1/2-i sqrt3/2)(x+1/2+ isqrt3/2)$

The expanded form can be directly obtained from

$(x^6-1)=prod(x-e^(i(2kpi)/6))$ , k=0, 1, 2,..,5.,

using the six 6th roots of 1, ${e^(i(2kpi)/6)}$ , k=0, 1, 2,..,5.,

How do you factor completely x^6-1x^6-1?